steam at 100 degree celcius is passed into a container of negligible heat capacity containing 20g of ice and 100g of water at 0 degree celcius untill the ice is evenly melted determine the total mass of water in the container(s.L of steam=2300,S.L of ice=340, Ch of water=4.2)
S.L=specific latent heat, Ch=specific heat capacity.
Total mass of water in the container = 120g.
The total heat energy required to melt the ice = 340 x 20 = 6800 J.
The total heat energy supplied by the steam = 2300 x 100 = 230000 J.
The remaining heat energy = 230000 - 6800 = 223200 J.
This heat energy is used to heat the 100g of water from 0°C to 100°C.
Heat energy required to heat the water = 100 x 4.2 x 100 = 420000 J.
Therefore, the total mass of water in the container = 100 + (223200/420000) x 100 = 120g.